No Trump Order by the Dealer?

[Richardb02]

Up Score: 0-0 Average & Equal Opponents (Actually weaker)
S4 (Dealer) R1
Would you pick up?

[RedDuke]

Probably not. You have the two aces, which is nice but the right+2 aces is only good for one trick in most cases. That Ace-Clubs is almost guaranteed to be trumped because you have the king too.

If none of your opponents ordered you up, I can almost guarantee that they're going to be loaded to the teeth with diamonds. If they had nothing, they'd have ordered you up just to stop a loner.

[Wes (aka the legend)]

Up Score: 0-0 Average & Equal Opponents (Actually weaker)
S4 (Dealer) R1
Would you pick up?

Right + two off aces when I block no suits is an auto-call for me. The bottom of my range when I block nothing is Right + a doubleton green ace or Right + a singleton Next ace. Having two off aces is a luxury!

[Wes (aka the legend)]

Probably not. You have the two aces, which is nice but the right+2 aces is only good for one trick in most cases. That Ace-Clubs is almost guaranteed to be trumped because you have the king too.

We have a guaranteed trick with the Right. I mean I don't have any hard data but surely a singleton Next Ace + a doubleton green ace will equate to at least 1 trick on average. So we got 2 tricks here ime.

If none of your opponents ordered you up, I can almost guarantee that they're going to be loaded to the teeth with diamonds. If they had nothing, they'd have ordered you up just to stop a loner.

This logic doesn't work becuz the vast majority of people do not donate, and even players that do donate typically don't do it at 0-0 (a strategy adjustment I disagree with but is very common amongst donaters).

This spot is marginal but it's not that bad. The dealer has 2 tricks on average and the 2 seat, his partner will still have 2 trump or 1 trump + the remaining off ace over 50% of the time.

The math behind my claim:

Probability S2 has two trump: (6C2 x 12C3)/18C5 = 38.52%
Probability S2 has 1 trump + AS: (6C1 x 1C1 x 11C3)/18C5 = 11.55%

Add the probabilities together and you get 50.07%

Edit: And that's an underestimate becuz the total range of possible hands (18C5 = 8,568) is not accurate becuz there are 1,506 hand combos of 3 trump, 4 trump, and 5 trump hands S2 can never have given that he passed. So the true probability is more like 60.75% when we adjust for the fact that S2's total possible hand combo range is (8,568 - 1,506 = 7,062).

The main assumption I'm working with here is S2 will never order up the dealer with 2 trump or less which I know isn't always true but I do think it's a good approximation.

Also keep in mind the cost of passing when you block no suits is higher than usual which means the dealer can get euchred more often and still turn a profit so to speak.

[Richardb02]

Up Score: 0-0 Average & Equal Opponents (Actually weaker)
S4 (Dealer) R1
Would you pick up?

With 2 off suit aces I expected to take a trick and lead the Right. BPS suggested a 70-75% chance of success.At 0-0 (any neutral score < 3-3)I like to experiment. With weaker competition, I expect poor decisions, not necessarily less aggressive decisions. So I went with my expectation of success at 70-75% (my minimum is 65%, basically break-even). So:
I ordered. I discarded the
Play went:
____T1____T2
S1
S2
S3
S4
We got 2 points!!